What is the derivative of Axe^{x/2}+B ?

The derivative of Axe^{x/2}+B is A((e^{x/2}x)/2+e^{x/2})

What is the first derivative of Axe^{x/2}+B ?

The first derivative of Axe^{x/2}+B is A((e^{x/2}x)/2+e^{x/2})

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\frac{d}{d x} (A x e^{\frac{x}{2}} + B)

A (\frac{e ^{\frac{x}{2}} x}{2} + e^{\frac{x}{2}})

Solution steps

\frac{d}{d x} (A x e^{\frac{x}{2}} + B)

Treat

A, B

as constants

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (A x e^{\frac{x}{2}}) + \frac{d B}{d x}

\frac{d}{d x} (A x e^{\frac{x}{2}}) = A (e^{\frac{x}{2}} + \frac{e ^{\frac{x}{2}} x}{2})

\frac{d B}{d x} = 0

= A (e^{\frac{x}{2}} + \frac{e ^{\frac{x}{2}} x}{2}) + 0

Simplify

= A (\frac{e ^{\frac{x}{2}} x}{2} + e^{\frac{x}{2}})

What is the derivative of Axe^{x/2}+B ?
The derivative of Axe^{x/2}+B is A((e^{x/2}x)/2+e^{x/2})
What is the first derivative of Axe^{x/2}+B ?
The first derivative of Axe^{x/2}+B is A((e^{x/2}x)/2+e^{x/2})